Let's solve the series step-by-step: Given series: \[ \sum_{n=0}^{\infty} \frac{(2x-3)^{2n+1}}{\left(\frac{3}{2}\right)^n} \] ### **Step 1: Rewrite the General Term** \[ a_n = \frac{(2x-3)^{2n+1}}{\left(\frac{3}{2}\right)^n} = (2x-3)^{2n+1} \cdot \left(\frac{2}{3}\right)^n = (2x-3) \cdot \left[(2x-3)^2 \cdot \frac{2}{3}\right]^n \] ### **Step 2: Identify as a Power Series** Let \( y = (2x-3)^2 \cdot \frac{2}{3} \), so the series is: \[ \sum_{n=0}^{\infty} (2x-3) \cdot y^n \] But since the first power is odd, let's focus on the root or ratio test for the original form. ### **Step 3: Use the Ratio Test to Find Radius and Interval of Convergence** Let’s use the root test (easier for powers): \[ \lim_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty} \sqrt[n]{|(2x-3)^{2n+1}|} \cdot \sqrt[n]{\left(\frac{2}{3}\right)^n} = |2x-3|^2 \cdot \frac{2}{3} \] For convergence, we need: \[ |2x-3|^2 \cdot \frac{2}{3} < 1 \] \[ |2x-3|^2 < \frac{3}{2} \] \[ |2x-3| < \sqrt{\frac{3}{2}} \] \[ -\sqrt{\frac{3}{2}} < 2x-3 < \sqrt{\frac{3}{2}} \] So: \[ 3 - \sqrt{\frac{3}{2}} < 2x < 3 + \sqrt{\frac{3}{2}} \] \[ \frac{3 - \sqrt{\frac{3}{2}}}{2} < x < \frac{3 + \sqrt{\frac{3}{2}}}{2} \] **Interval of convergence:** \[ \left( \frac{3 - \sqrt{3/2}}{2}, \frac{3 + \sqrt{3/2}}{2} \right) \] **Radius of convergence:** \[ R = \frac{\sqrt{3/2}}{2} \] --- ### **Step 4: Absolute and Conditional Convergence** The series converges absolutely **inside** the interval (excluding endpoints). To check for **conditional** convergence, check the endpoints: #### **Check \( x_1 = \frac{3 - \sqrt{3/2}}{2} \):** At this point, \( |2x-3| = \sqrt{3/2} \), so the general term becomes: \[ a_n = (\pm \sqrt{3/2})^{2n+1} \cdot \left(\frac{2}{3}\right)^n = (\pm 1)^{2n+1} \cdot (\sqrt{3/2}) \cdot 1^n \] But \( (\pm 1)^{2n+1} = \pm 1 \), so \( a_n \) does **not tend to zero**; it oscillates and doesn't diminish. **So it diverges at endpoints.** #### **Check \( x_2 = \frac{3 + \sqrt{3/2}}{2} \):** Similar reasoning as above, the term does not go to zero. **So, the series does not converge conditionally at any point; it only converges absolutely inside the interval.** --- ## **Final Answers:** ### (a) The interval of convergence is: \[ \left( \frac{3 - \sqrt{3/2}}{2}, \frac{3 + \sqrt{3/2}}{2} \right) \] The radius of convergence is: \[ \frac{\sqrt{3/2}}{2} \] ### (b) The series converges absolutely on: \[ \left( \frac{3 - \sqrt{3/2}}{2}, \frac{3 + \sqrt{3/2}}{2} \right) \] ### (c) For what values of \( x \) does the series converge conditionally? \[ \boxed{\text{B. The series does not converge conditionally.}} \] --- **Boxed summary for direct input:** - (a) Interval of convergence: \(\left(\frac{3-\sqrt{3/2}}{2}, \frac{3+\sqrt{3/2}}{2}\right)\) - Radius of convergence: \(\frac{\sqrt{3/2}}{2}\) - (b) Absolutely: \(\left(\frac{3-\sqrt{3/2}}{2}, \frac{3+\sqrt{3/2}}{2}\right)\) - (c) **B. The series does not converge conditionally.**